A description/proof of that for vectors bundle, trivializing open subset is not necessarily chart open subset, but there is possibly smaller chart trivializing open subset at any point on trivializing open subset
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) vectors bundle.
- The reader admits the proposition that the trivialization of any chart trivializing open subset is a chart map.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at any point on any trivializing open subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any \(C^\infty\) manifold, \(M\), any vectors bundle, \(\pi: E \rightarrow M\), and any trivializing open subset, \(U\), \(U\) is not necessarily any chart open subset, but around any point, \(p \in U\), there is a possibly smaller chart trivializing open subset, \(U'_p \subseteq U\).
2: Proof
\(U\) is not necessarily a chart open subset, because \(U\)'s being a trivializing open set does not guarantee that \(U\) is a chart open set. For example, for the product bundle, \(M \times \mathbb{R}^d\), \(U = M\) is a trivializing open set, but \(M\) does not necessarily have the global chart, so, \(U\) is not necessarily a chart open set.
Around \(p\), there is a chart open subset, \(U_p \subseteq M\). \(U'_p := U_p \cap U\) is a chart open subset. While there is a trivialization, \(\phi: \pi^{-1} (U) \rightarrow U \times \mathbb{R}^d\), \(\phi\vert_{\pi^{-1} (U'_p)}: \pi^{-1} (U'_p) \rightarrow U'_p \times \mathbb{R}^d\) is a trivialization, because it is fiber-preserving, diffeomorphic, and for any \(p' \in U'_p\), \(\phi\vert_{\pi^{-1} (U'_p)}\vert_{\pi^{-1} (p')}: \pi^{-1} (p') \rightarrow \{p'\} \times \mathbb{R}^d\) is vectors space isomorphic, because it is just a restriction of \(\phi\) on an open subset domain (\(\pi^{-1} (U'_p)\) is open on \(E\), and \(\pi^{-1} (U'_p) \subseteq \pi^{-1} (U)\), which is open on \(E\)).
3: Note
If \(U\) is not any chart open subset on \(M\), \(\pi^{-1} (U)\) is not necessarily a chart open subset on \(E\), while \(\pi^{-1} (U'_p)\) is a chart open subset on \(E\), by the proposition that the trivialization of any chart trivializing open subset is a chart map.
